Given a manifold, we can generalize the idea of derivatives in multiple ways: two of them being the Lie derivative and the covariant derivative. Whereas Lie derivatives do not require any additional structure to be defined on a manifold, covariant derivatives need connections to be well-defined. Also, Lie derivatives are used to define symmetries of a tensor field whereas covariant derivatives are used to define parallel transport.

The symmetries of a tensor T are given by the one-parameter family of diffeomorphisms $\phi_t$ generated by a vector field $V$ which satisfies $\mathcal{L}_V T=0$, or $\phi_t^*T=T$.

A curve generated by a vector field X keeps a tensor T unchanged under parallel transport obeys $\nabla_X T = 0$.

My question is the following, why do we need covariant derivative which requires additional structure to find curves along which a tensor is parallel-transported. Since Lie derivatives give that along $\phi_t$ the tensor remains unchanged (and hence parallel to itself) if $\mathcal{L}_V T=0$. So, if we define the vector field X (in the case of covariant derivative) to be V, we have a curve which parallel transports the tensor and keeps it unchanged. So, why do we need to solve geodesic equations or equations of parallel transport when we can just solve the Lie derivative equation? What extra information does the covariant derivative give us in this case?

P.S. I have read the following two PhysicsSE posts which discuss some aspects of it but does not address this question per se.

  1. Lie derivative vs. covariant derivative in the context of Killing vectors

  2. Why do we need connections, if we have the Lie derivative?

  • 1
    $\begingroup$ Maybe I am wrong, but the Lie derivative doesn't recquire a notion of distance as you have to take a limit? In general with differentiable manifold you have the topology and the atlas $\endgroup$
    – Ratman
    Commented Jul 23, 2020 at 23:26
  • $\begingroup$ Maybe I am missing something, but how is this related to the question? That just says that Lie derivative is a more primitive notion of a derivative and covariant derivative requires more structure to be defined. $\endgroup$
    – abhijit975
    Commented Jul 24, 2020 at 1:02
  • 1
    $\begingroup$ It was just a comment about your first paragraph, in the Lie derivative case you need a notion of distance to define it, maybe a metric. So It recquires more structure than what a diff manifold normally has. I am completely aware this doesn't solve your problema, It Is just an observation $\endgroup$
    – Ratman
    Commented Jul 24, 2020 at 8:06

2 Answers 2


Consider the expression $\nabla_{\mathbf X_p} \mathbf T$, for some arbitrary tangent vector $\mathbf X_p$ and tensor field $\mathbf T$. Since a tangent vector is always defined at a point $p$, we are to understand this expression to mean the rate of change of $\mathbf T$ along $\mathbf X_p$ at the point $p$ (where $\mathbf X_p$ lives). If $\mathbf T$ is a $(r,s)$-tensor field, then the quantity $\nabla_{\mathbf{X}_p} \mathbf T$ is a $(r,s)$-tensor at the point $p$.

Of course, if you have a vector field $\mathbf Y$, then we can define the object $\nabla_{\mathbf Y}\mathbf T$ to be a $(r,s)$-tensor field. The tensor at the point $p$ is then simply $\left(\nabla_\mathbf Y \mathbf T\right)_p := \nabla_{\mathbf Y_p}\mathbf T$.

The point is that the covariant derivative is taken with respect to a vector, not a vector field, and returns a $(r,s)$-tensor, not an $(r,s)$-tensor field. If we happen to have a vector field lying around then we can of course plug in the corresponding vector at every point to obtain a $(r,s)$-tensor field, but this is not necessary. Contrast this with the Lie derivative, for which $\mathcal L_\mathbf X \mathbf T$ requires $\mathbf X$ to be a vector field. This is the price we pay for not introducing the additional structure of a connection; the Lie derivative requires information about $\mathbf X$ in a neighborhood, not just at a point.

My question is the following, why do we need covariant derivative which requires additional structure to find curves along which a tensor is parallel-transported.

If I give you a curve $\gamma$ and ask you to parallel-transport a tensor $\mathbf T$ along it with the Lie derivative, then what would you do? Presumably you'd compute the tangent vector $\mathbf V_p= \gamma'(p)$ at every point $p$ along the curve, and then plug it into $\mathcal L_{\mathbf V_p} \mathbf T=0$. However, in computing this, you'd find yourself taking the derivatives of the components of $\mathbf V_p$ in arbitrary directions, not just along $\gamma$.

How would you do this? If all you have is the curve, then it doesn't even make sense to ask how the tangent vector changes in arbitrary directions. You'd need to define a vector field in the neighborhood of $\gamma$ for which $\gamma$ is an integral curve, and herein lies the problem: there are an infinity of vector fields which fit this description, and parallel-transporting via two different such vector fields would generically give two different results.

In order for this to be a well-defined procedure, you need additional structure - namely, a choice of vector field (for which $\gamma$ is an integral curve) along which to parallel-transport your tensor. If you want this procedure to be consistent for all curves, then you'd need to provide some kind of criterion for choosing the "correct" vector field given only one of its integral curves, and this is precisely the additional structure which is required to define a connection.

  • $\begingroup$ Shouldn’t the covariant derivative of a $(r,s)$ tensor field be of type $(r,s+1)$? $\endgroup$
    – Bcpicao
    Commented Feb 28 at 19:00
  • $\begingroup$ @Bcpicao I am talking about $\nabla_{\mathbf X_p}\mathbf Y$, the covariant derivative of the vector field $\mathbf Y$ along the vector $\mathbf X_p$ defined at the point $p$. You are talking about $\nabla_\cdot \mathbf Y \equiv \nabla \mathbf Y$, which is essentially the same object in which you've left the bottom slot open. There is some subtlety here related to the vector vs. vector field point that forms the crux of this answer, so it's worth being a bit careful. $\endgroup$
    – J. Murray
    Commented Feb 29 at 12:47

I'm not sure but I think one big difference is that a covariant derivative defines a general way to parallel transport vectors along arbitrary curves, giving a solution that only depends on that curve, whereas the Lie derivative always depends on some vector field X and thus could only be used to transport vectors along its integral curves. Thus if you wanted transport along any curves you would need to use different vector fields for different curves. But this transport depends not only on the values of X on that curve but in a whole neighbourhood. Thereforre such a definition would not be well defined and is therefore useless. So basically I think, there is no way to define a parallel transport along arbitrary curves using Lie derivatives.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.